English
Karnataka Board PUCPUC Science Class 11

A 10 Kw Drilling Machine is Used to Drill a Bore in a Small Aluminium Block of Mass 8.0 Kg. How Much is the Rise in Temperature of the Block in 2.5 Minutes, Assuming 50% of Power is Used up in Heating the Machine Itself Or Lost to the Surroundings

Advertisements
Advertisements

Question

A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings Specific heat of aluminium = 0.91 J g–1 K–1

Advertisements

Solution 1

Power of the drilling machine, P = 10 kW = 10 × 10W

Mass of the aluminum block, m = 8.0 kg = 8 × 103 g

Time for which the machine is used, t = 2.5 min = 2.5 × 60 = 150 s

Specific heat of aluminium, c = 0.91 J g–1 K–1

Rise in the temperature of the block after drilling = δT

Total energy of the drilling machine = Pt

= 10 × 10× 150

= 1.5 × 106 J

It is given that only 50% of the power is useful.

Useful energy, `triangle Q = 50/100 xx 1.5 xx 10^6 = 7.5xx10^5 J`

But `triangle Q = mctriangle T`

`:. triangle T =  (triangle Q)/"mc"`

`= (7.5 xx 10^5)/(8xx10^3xx0.91)`

`= 103 ^@C`

Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is 103°C.

shaalaa.com

Solution 2

Power = 10 kW = 104 W

Mass, m=8.0 kg = 8 x 103 g

Rise in temperature, `triangle T =?`

`Time, t = 2.5 min = 2.5 xx 60 = 150 s`

Specific heat, `C = 0.91 Jg^(-1) K^(-1)`

Total energy =  Power x Time  = `10^4 xx 150 J`

`=15 xx 10^5 J`

As 50% of energy is lost

∴Thermal energy available

`triangle Q =  1/2 xx 15 xx 10^5 = 7.5 xx 10^5 J`

Since `triangle Q = mctrangle T`

`:.triangle T = triangleQ/mc = (7.5xx20^5)/(8xx10^3xx0.91) = 103^@C`

shaalaa.com
  Is there an error in this question or solution?

RELATED QUESTIONS

A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10–5 K–1.


A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1, steel = 1.2 × 10–5 K–1).


If two bodies are in thermal equilibrium in one frame, will they be in thermal equilibrium in all frames?


The density of water at 0°C is 0.998 g cm–3 and at 4°C is 1.000 g cm–1. Calculate the average coefficient of volume expansion of water in the temperature range of 0 to 4°C.


Show that the moment of inertia of a solid body of any shape changes with temperature as I = I0 (1 + 2αθ), where I0 is the moment of inertia at 0°C and α is the coefficient of linear expansion of the solid.


Answer the following question.

Derive the relation between three coefficients of thermal expansion.


Solve the following problem.

A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the wooden rim and the iron ring are 1.5 m and 1.47 m respectively at room temperature of 27 °C. To what temperature the iron ring should be heated so that it can fit the rim of the wheel? (αiron = 1.2 × 10–5K–1).


A clock pendulum having coefficient of linear expansion. α = 9 × 10-7/°C-1 has a period of 0.5 s at 20°C. If the clock is used in a climate, where the temperature is 30°C, how much time does the clock lose in each oscillation? (g = constant)


A metal sphere 10.01 cm in diameter is placed on a brass ring of internal diameter 10 cm and at the same temperature of 12° C. The temperature up to which they should be heated together so that the metal sphere just passes through the ring is `[alpha_"metal"= 12 xx 10^-6//°"C" and alpha_"brass" =18 xx 10^-6//°"C"]` ____________.


A bimetallic strip is made of aluminium and steel (αAl > αsteel) . On heating, the strip will ______.


A student records the initial length l, change in temperature ∆T and change in length ∆l of a rod as follows:

S.No. l(m) ∆T (C) ∆l (m)
1. 2 10 `4 xx 10^-4`
2. 1 10 `4 xx 10^-4`
3. 2 20 `2 xx 10^-4`
4. 3 10 `6 xx 10^-4`

If the first observation is correct, what can you say about observations 2, 3 and 4.


Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of 57°C is drunk. You can take body (tooth) temperature to be 37°C and α = 1.7 × 10–5/°C, bulk modulus for copper = 140 × 109 N/m2.


The height of mercury column measured with brass scale at temperature T0 is H0. What height H' will the mercury column have at T = 0°C. Coefficient of volume expansion of mercury is γ. Coefficient of linear expansion of brass is α ______.


Length of steel rod so that it is 5 cm longer than the copper rod at all temperatures should be ______ cm.

(α for copper = 1.7 × 10-5/°C and α for steel = 1.1 × 10-5/°C)


A disc is rotating freely about its axis. The percentage change in angular velocity of a disc if temperature decreases by 20°C is ______.

(coefficient of linear expansion of material of disc is 5 × 10-4/°C)


A metal rod Y = 2 × 1012 dyne cm-2 of coefficient of linear expansion 1.6 × 10-5 per °C has its temperature raised by 20°C. The linear compressive stress to prevent the expansion of the rod is ______.


The increase in the dimensions of a body due to an increase in its temperature is called ______.


When a solid is heated, its atoms vibrate faster and move farther apart. This happens because ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×